POJ - 2411 Mondriaan's Dream （状压DP）

Mondriaan's Dream

 

Squares and rectangles fascinated the famous Dutch painter Piet Mondriaan. One night, after producing the drawings in his 'toilet series' (where he had to use his toilet paper to draw on, for all of his paper was filled with squares and rectangles), he dreamt of filling a large rectangle with small rectangles of width 2 and height 1 in varying ways. 

Expert as he was in this material, he saw at a glance that he'll need a computer to calculate the number of ways to fill the large rectangle whose dimensions were integer values, as well. Help him, so that his dream won't turn into a nightmare!

Input

The input contains several test cases. Each test case is made up of two integer numbers: the height h and the width w of the large rectangle. Input is terminated by h=w=0. Otherwise, 1<=h,w<=11.

Output

For each test case, output the number of different ways the given rectangle can be filled with small rectangles of size 2 times 1. Assume the given large rectangle is oriented, i.e. count symmetrical tilings multiple times.

Sample Input

1 2
1 3
1 4
2 2
2 3
2 4
2 11
4 11
0 0

Sample Output

1
0
1
2
3
5
144
51205

题目链接：http://poj.org/problem?id=2411

题目大意：问n*m的矩形用1*2的小矩形填充，问填满一共有多少种方案

思路：如果n和m都是奇数，那么n*m也是奇数无法填满，方案数都为0。

dp[ i ] [ j ] 表示 前 i-1行填满 且 第 i 行的状态为 j 的方案数

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll long long
const int N=(1<<11)+5;
ll dp[12][N],ans[12][12];
int n,m,ma;
int init(int x)
{
    int cnt=0;
    for(int i=0;i<m;)
    {
        if(x&(1<<i))
        {
            if(i==m-1) return 0;
            if(x&(1<<(i+1))) i+=2;
            else return 0;
        }
        else i++;
    }
    return 1;
}
int fun(int x,int y)
{
    for(int i=0;i<m;)
    {
        if(x&(1<<i))
        {
            if(y&(1<<i))
            {
                if(i==m-1) return 0;
                if((x&1<<(i+1))==0||(y&1<<(i+1))==0) return 0;
                i+=2;
            }
            else i++;
        }
        else
        {
            if(y&(1<<i))i++;
            else return 0;
        }
    }
    return 1;
}
int main()
{
    memset(ans,-1,sizeof(ans));
    while(~scanf("%d%d",&n,&m)&&(n+m))
    {
        if(n%2&&m%2) ans[n][m]=ans[m][n]=0;
        if(ans[n][m]>-1) printf("%lld\n",ans[n][m]);
        else
        {
            if(n<m) swap(n,m);
            ma=1<<m;
            memset(dp,0,sizeof(dp));
            for(int i=0;i<ma;i++)
                if(init(i)) dp[1][i]=1;
            for(int i=2;i<=n;i++)
            {
                for(int j=0;j<ma;j++)
                    for(int k=0;k<ma;k++)
                    if(fun(j,k)) dp[i][k]+=dp[i-1][j];
            }
            ans[n][m]=ans[m][n]=dp[n][ma-1];
            printf("%lld\n",dp[n][ma-1]);
        }
    }
    return 0;
}